Voltage gain for amplifier (JFET) in common gate

[adrms]

Homework Statement

Prove that the voltage gain for the following amplifier in common gate is $$A_V=\frac{(g_mr_d+1)R_L}{R_S(g_mr_d+1)+R_L+r_d}$$

Relevant Equations

* We use the Hybrid model to analyze the transistor.

$$A_V=\frac{(g_mr_d+1)R_L}{R_S(g_mr_d+1)+R_L+r_d}$$

First the hybrid model, I assume the capacitor works as a short circuit regarding the altern current:

$$A_V=\frac{v_o}{v_i}$$
$$v_i=-v_{GS}$$
$$v_o=-i_LR_L$$
$$i_L=g_mv_{GS}+\frac{v_{DS}}{r_d}$$

Now I use the Kirchhoff's law to get $v_{DS}$. I consider this close loop:

$$-v_i-v_{DS}+v_o=0$$
$$v_{DS}=-v_i+v_o=v_{GS}+v_o$$
$$i_L=g_mv_{GS}+\frac{v_{GS}+v_o}{r_d}$$
$$v_o=-i_LR_L=-\left(g_mv_{GS}+\frac{v_{GS}+v_o}{r_d}\right)R_L$$
Now I just try to get $v_o$ from that equation:
$$v_o=-g_mv_{GS}R_L-\frac{v_{GS}+v_o}{r_d}R_L$$
$$v_o=-g_mv_{GS}R_L-\frac{v_{GS}}{r_d}R_L-\frac{v_o}{r_d}R_L$$
$$v_o+\frac{v_o}{r_d}R_L=-g_mv_{GS}R_L-\frac{v_{GS}}{r_d}R_L$$
$$v_o\left(1+\frac{R_L}{r_d}\right)=-g_mv_{GS}R_L-\frac{v_{GS}}{r_d}R_L$$
$$v_o=\frac{-g_mv_{GS}R_L-\frac{v_{GS}}{r_d}R_L}{1+\frac{R_L}{r_d}}$$
$$v_o=\frac{-v_{GS}R_L\left(g_m+\frac{1}{r_d}\right)}{1+\frac{R_L}{r_d}}$$
$$v_o=\frac{v_iR_L\left(g_m+\frac{1}{r_d}\right)}{1+\frac{R_L}{r_d}}$$
$$A_V=\frac{v_o}{v_i}=\frac{R_L\left(g_m+\frac{1}{r_d}\right)}{1+\frac{R_L}{r_d}}\cdot\frac{r_d}{r_d}$$
$$\boxed{A_V=\frac{R_L\left(g_mr_d+1\right)}{r_d+R_L}}$$

The answer given by my teacher is:
$$A_V=\frac{(g_mr_d+1)R_L}{R_S(g_mr_d+1)+R_L+r_d}$$

As you can see I am missing a whole term in the denominator. I do not understand how $R_S$ gets into the equation.

 

[Baluncore]

I expect your assumption about the input capacitor may make the difference.
https://courses.cit.cornell.edu/ece315/Lectures/handout12.pdf
https://en.wikipedia.org/wiki/Common_gate#Low-frequency_characteristics